Calculate the pressure of the following mixture.

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SUMMARY

The discussion focuses on calculating the pressure at various depths in an oil reservoir using the equation P = rho g h + Patm. The specific densities provided are 1000 kg/m³ for water, 800 kg/m³ for oil, and 120 kg/m³ for gas. The calculated pressure at the water/oil contact (WOC) is 64.7 MPa, with further calculations for depths of 12500 ft, 11700 ft, and 11000 ft yielding pressures of 34.7 MPa, 67.1 MPa, and 98.2 MPa, respectively. The participants emphasize the importance of using the correct height and density values for accurate pressure calculations.

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Homework Statement



An oil reservoir contains at water/oil contact (WOC) of 12000ft contains a water column at 800ft, oil at 500ft and gas at 600ft height

Calculate the pressure at WOC, 12500, 11700 and 11000 ft?

Homework Equations



P = rho g h + Patm

1 ft = 0.305m
Rho water = 1000
Rho Oil = 800
Rho gas = 120

The Attempt at a Solution



P WOC = 1000*9.81*(12000*0.305) + 800*9.81*(12000*0.305)+patm = 64.7Mpa

For the water column for example would the pressure be:

Pwoc + (1000*9.81*(800*0.305) = 67.1Mpa ?
or
Pwoc + (1000*9.81*((12000-800)*0.305)) = 98.2Mpa
or
Pwoc+(1000*9.81*((12000+800)*0.305)) = 102.3Mpa

Basically what height do I have to use?

And what densitys do I need to use for the depths at 12000, 11700 and 11000ft please?
 
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Pwoc = (rhoWgh ) + (rhoOgh)
= (1000*9.8*(12000*0.305)) + (800*9.8*(12000*0.305)
= 64.6 Mpa

Pgoc = (rhoG gh) + (rhoO gh)
= ((120*9.8*((12000-800)*0.305))) + (800*9.8*((12000-800)*0.305)))
= 30.8 Mpa

P12500 = (rhoW gh)
= 1000*9.8*(12500*0.305)
= 34.7 Mpa
 

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